The Integral Test | Real Analysis

A series and an integral get stacked together and share one fate. The bars wrap the curve from above and below.

Stand bars under a decreasing curve, and the sum traps the integral from both sides. So the series and the integral converge or diverge together. An exponent above 1 gives convergence; anything else diverges. That fork decides the fate of every p-series.

For example, stack 📚 books like steps against a ramp. The steps dip below the ramp's curve and poke above it. Still, the steps' area and the area under the curve are finite together or endless together.

Examples & Exercises

Prove the integral test completely: if $f$ is positive and decreasing on $[1,\infty)$ , then on each $[n,n+1]$ we have $f(n+1) \le \int_n^{n+1} f \le f(n)$ ; summing gives $\int_1^{N+1} f \le \sum_{n=1}^{N} f(n) \le f(1) + \int_1^{N+1} f$ , so $\sum f(n)$ converges iff $\int_1^{\infty} f$ converges.

f(n+1) \le \int_n^{n+1} f \le f(n) \qquad (f \text{ decreasing})

\int_1^{N+1} f \le \sum_{n=1}^{N} f(n) \le f(1) + \int_1^{N+1} f

p = 2, N = 12

\int_1^{13} x^{-2}\,dx = 12/13 \approx 0.9231 \le \sum_{k=1}^{12} k^{-2} \approx 1.5650 \le 1 + 0.9231

— the sandwich, numerically.

Show proof

First, the integrals exist: a monotone function on $[a, b]$ is Riemann integrable (standard theorem), so the decreasing $f$ is integrable on every $[n, n+1]$ and every $\int_n^{n+1} f$ exists.
Since $f$ is decreasing, $f(n+1) \le f(x) \le f(n)$ for $x \in [n, n+1]$ .
Integrating over $[n,n+1]$ (length $1$ ) gives $f(n+1) \le \int_n^{n+1} f \le f(n)$ by monotonicity of the integral.
Sum over $n = 1, \ldots, N$ .
The left inequalities sum to $\sum_{n=1}^{N} f(n+1) = \sum_{k=2}^{N+1} f(k) \le \int_1^{N+1} f$ ; the right ones to $\int_1^{N+1} f \le \sum_{n=1}^{N} f(n)$ .
With $S_N = \sum_{n=1}^{N} f(n)$ , the correct identity is $\sum_{k=2}^{N+1} f(k) = S_{N+1} - f(1) = S_N - f(1) + f(N+1)$ .
Hence the left side gives $S_{N+1} - f(1) \le \int_1^{N+1} f$ , i.e. $S_{N+1} \le f(1) + \int_1^{N+1} f$ , and since $f > 0$ , $S_N \le S_{N+1}$ , so $S_N \le f(1) + \int_1^{N+1} f$ as well.
Chaining with the right side gives $\int_1^{N+1} f \le \sum_{n=1}^{N} f(n) \le f(1) + \int_1^{N+1} f$ .
$(\Leftarrow)$ If $\int_1^{\infty} f$ converges (i.e. $\int_1^{R} f$ has a finite limit as $R \to \infty$ ), then the partial sums $\sum_{n=1}^{N} f(n)$ are bounded above by $f(1) + \int_1^{\infty} f$ and, the terms being nonnegative, converge by the monotone convergence theorem.
$(\Rightarrow)$ By contraposition: if $\int_1^{\infty} f$ diverges (to $+\infty$ , the integrand being positive), then $\sum_{n=1}^{N} f(n) \ge \int_1^{N+1} f \to \infty$ , so the series diverges too.
Hence $\sum f(n)$ converges $\iff$ $\int_1^{\infty} f$ converges.
QED.

Prove the $p$ -series $\sum_{n=1}^{\infty} n^{-p}$ converges for $p > 1$ and diverges for $p \le 1$ , completely: compute $\int_1^{R} x^{-p}\,dx$ directly ( $p = 1$ gives $\ln R$ ) and apply the integral test.

\int_1^{R} x^{-p}\,dx = \begin{cases} \ln R & p = 1 \\ \dfrac{R^{1-p} - 1}{1-p} & p \ne 1 \end{cases}

\sum_{n=1}^{\infty} n^{-p} \text{ converges } \iff p > 1

p = 1, N = 12

: the harmonic partial sum

H_{12} \approx 3.1032

keeps growing like

\ln

, and

\int_1^{13} \frac{dx}{x} = \ln 13 \approx 2.5649

— both diverge.

Show proof

First, $p \le 0$ : $n^{-p} \ge 1$ , so the general term does not tend to $0$ and the series diverges by the term test (on this range $f(x) = x^{-p}$ is increasing, so the integral test does not apply).
Now assume $p > 0$ ; then $f(x) = x^{-p}$ is positive and decreasing on $[1,\infty)$ , so the integral test applies.
Compute: for $p = 1$ , $\int_1^{R} x^{-1}\,dx = \ln R$ ; for $p \ne 1$ , $\int_1^{R} x^{-p}\,dx = \left[\frac{x^{1-p}}{1-p}\right]_1^{R} = \frac{R^{1-p} - 1}{1-p}$ .
$p > 1$ : $1 - p < 0$ , so $R^{1-p} \to 0$ as $R \to \infty$ . Hence $\int_1^{\infty} x^{-p}\,dx = \frac{1}{p-1} < \infty$ — the integral converges, so by the integral test $\sum n^{-p}$ converges.
$p = 1$ : $\ln R \to \infty$ , so the integral diverges, and hence the harmonic series $\sum 1/n$ diverges. $0 < p < 1$ : $1 - p > 0$ , so $\frac{R^{1-p}-1}{1-p} \to \infty$ — the integral diverges and, by the integral test, so does the series (the case $p \le 0$ was handled by the term test in step 1).
In summary: $\sum n^{-p}$ converges $\iff$ $\int_1^{\infty} x^{-p}\,dx < \infty$ $\iff$ $p > 1$ . QED.

Prove the remainder estimate completely: if $f$ is positive, decreasing, and $\sum f(n)$ converges, the tail $R_N = \sum_{n>N} f(n)$ satisfies $\int_{N+1}^{\infty} f \le R_N \le \int_{N}^{\infty} f$ .

Numerically: trap the $N = 10$ tail of $\sum 1/n^2$ between $1/11 \approx 0.0909$ and $1/10 = 0.1$ .

\int_{N+1}^{\infty} f \le R_N = \sum_{n>N} f(n) \le \int_{N}^{\infty} f

\frac{1}{11} \approx 0.0909 \le R_{10}\!\left(\sum \tfrac{1}{n^2}\right) \le 0.1 = \frac{1}{10}

p = 2, N = 10

: the tail

R_{10}

lies between

\int_{11}^{\infty} x^{-2}\,dx = 1/11 \approx 0.0909

and

\int_{10}^{\infty} x^{-2}\,dx = 1/10 = 0.1

(true value

\approx 0.09517

Show proof

Sum the basic inequality $f(n+1) \le \int_n^{n+1} f \le f(n)$ (first problem) over $n = N+1, \ldots, M$ .
Left: $\sum_{n=N+1}^{M} f(n+1) = \sum_{k=N+2}^{M+1} f(k) \le \int_{N+1}^{M+1} f$ .
Right: $\int_{N+1}^{M+1} f \le \sum_{n=N+1}^{M} f(n)$ .
A shifted version of the right inequality: the basic inequality on $[n-1,n]$ reads $f(n) \le \int_{n-1}^{n} f$ , so summing over $n = N+1,\ldots,M$ gives $\sum_{n=N+1}^{M} f(n) \le \int_{N}^{M} f$ .
Let $M \to \infty$ .
Since the series converges, $\sum_{n=N+1}^{M} f(n) \to R_N$ (finite), and by the integral test the improper integrals converge too, so $\int_{N+1}^{M+1} f \to \int_{N+1}^{\infty} f$ and $\int_{N}^{M} f \to \int_{N}^{\infty} f$ .
Passing to the limit preserves the inequalities: $\int_{N+1}^{\infty} f \le R_N \le \int_{N}^{\infty} f$ .
QED.
Numerical application: $f(x) = x^{-2}$ , $N = 10$ .
$\int_{11}^{\infty} x^{-2}\,dx = \left[-x^{-1}\right]_{11}^{\infty} = \frac{1}{11} \approx 0.0909$ and $\int_{10}^{\infty} x^{-2}\,dx = \frac{1}{10} = 0.1$ , so $\frac{1}{11} \le R_{10} \le \frac{1}{10}$ .
Indeed $R_{10} = \frac{\pi^2}{6} - \sum_{k=1}^{10}\frac{1}{k^2} \approx 0.09517$ lands inside the interval.